\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2010.3.1 Question 1

1. Notice that \begin {align*} C & = \frac {1}{n + 1} \sum _{k = 1}^{n + 1} x_k \\ & = \frac {1}{n + 1} \left (\sum _{k = 1}^{n} x_k + x_{n + 1}\right ) \\ & = \frac {1}{n + 1} \left (n A + x_{n + 1}\right ). \end {align*}
2. By expanding the brackets, \begin {align*} B & = \frac {1}{n} \sum _{k = 1}^{n} (x_k - A)^2 \\ & = \frac {1}{n} \sum _{k = 1}^{n} (x_k^2 - 2Ax_k + A^2) \\ & = \frac {1}{n} \left [\sum _{k = 1}^{n} x_k^2 - 2A \sum _{k = 1}^{n} x_k + A^2 n \right ] \\ & = \frac {1}{n} \sum _{k = 1}^{n} x_k^2 - 2A \frac {1}{n} \sum _{k = 1}^{n} x_k + A^2 \\ & = \frac {1}{n} \sum _{k = 1}^{n} x_k^2 - 2A^2 + A^2 \\ & = \frac {1}{n} \sum _{k = 1}^{n} x_k^2 - A^2. \end {align*}

3. Similarly, we have \[ D = \frac {1}{n + 1} \sum _{k = 1}^{n + 1} x_k^2 - C^2. \]

   Hence, \begin {align*} D & = \frac {1}{n + 1} \sum _{k = 1}^{n + 1} x_k^2 - C^2 \\ & = \frac {1}{n + 1} \left (\sum _{k = 1}^{n} x_k^2 + x_{n + 1}^2\right ) - \left (\frac {1}{n + 1} (nA + x_{n + 1})\right )^2 \\ & = \frac {1}{n + 1} \left (n (B + A^2) + x_{n + 1}^2\right ) - \left (\frac {1}{n + 1} (nA + x_{n + 1})\right )^2 \\ & = \frac {1}{(n + 1)^2} \left [(n + 1)\left (n (B + A^2) + x_{n + 1}^2\right ) - (nA + x_{n + 1})^2\right ] \\ & = \frac {1}{(n + 1)^2} \left (n A^2 + n(n + 1)B + n x_{n + 1}^2 - 2nA x_{n + 1}\right ) \\ & = \frac {n}{(n + 1)^2} \left (A^2 + (n + 1)B + x_{n + 1}^2 - 2A x_{n + 1}\right ) \\ & = \frac {n}{(n + 1)^2} \left [\left (A - x_{n + 1}\right )^2 + (n + 1)B\right ] \end {align*}

   Hence, \begin {align*} (n + 1)D - nB & = \frac {n}{n + 1} \left [\left (A - x_{n + 1}\right )^2 + (n + 1)B\right ] - nB \\ & = \frac {n}{n + 1} \cdot \left (A - x_{n + 1}\right )^2 + nB - nB \\ & = \frac {n}{n + 1} \cdot \left (A - x_{n + 1}\right )^2 \\ & \geq 0, \end {align*}

   since a square is always non-negative, and hence \[ (n + 1)D \geq nB. \]

   On the other hand, notice that \begin {align*} D - B & = \frac {n}{(n + 1)^2} \left [\left (A - x_{n + 1}\right )^2 + (n + 1)B\right ] - B \\ & = \frac {n}{(n + 1)^2} \left (A - x_{n + 1}\right )^2 + \frac {n}{n + 1} B - B \\ & =\frac {n}{(n + 1)^2} \left (A - x_{n + 1}\right )^2 - \frac {1}{n + 1}B, \end {align*}

   and hence \begin {align*} D < B & \iff \frac {n}{(n + 1)^2} \left (A - x_{n + 1}\right )^2 - \frac {1}{n + 1}B < 0 \\ & \iff \frac {n}{(n + 1)^2} \left (A - x_{n + 1}\right )^2 < \frac {1}{n + 1}B \\ & \iff \left (A - x_{n + 1}\right )^2 < \frac {n + 1}{n} B \\ & \iff - \sqrt {\frac {n + 1}{n} B} < A - x_{n + 1} < \sqrt {\frac {n + 1}{n} B} \\ & \iff - A - \sqrt {\frac {n + 1}{n} B} < - x_{n + 1} < - A + \sqrt {\frac {n + 1}{n} B} \\ & \iff A - \sqrt {\frac {n + 1}{n} B} < x_{n + 1} < A + \sqrt {\frac {n + 1}{n} B}, \end {align*}

   exactly as desired.